Minor Loss Coefficients 2

[Superfly257]

Hello All,

I was hoping to get some confirmation on some minor pressure loss calculations I am doing, seeing as how I am getting wildly different values depending on whether I try to use the minor loss coefficient, or the equivalent length method.

The minor loss system components are:
5 threaded, line flow tees
4 threaded, 90 degree elbows
2 fully open ball valves

I was using this reference from the Engineering Toolbox:

http://www.engineeringtoolbox.com/minor-loss-coefficients-pipes-d_626.html

I'm not sure if the correct way is simply to add the minor loss coefficients in order to get a total coefficient value, but that was the approach I used... I came up with a coefficient of 10.6.

I've attached a copy of the spreadsheet I am using for my calculations, so you can see all the values I am using.
Here is a copy of the link:

http://files.engineering.com/getfil...fdb436b169&file=Pressure_Loss_Calculator.xlsx

Any confirmation on either of these calculations, or maybe a helpful link with reliable information or a detailed example,would be very helpful.

Thanks in advance!
-Superfly

 

[Superfly257]

Also, I apologize in advance for all the seemingly random unit conversions...
I like doing my calculations in metric, but the outputs need to be in imperial units.

-Superfly

 

[ione]

If you deal with laminar flow (probably more than a guess, as I’ve seen your thread

http://www.eng-tips.com/viewthread.cfm?qid=292763&page=1)

you should be aware about more reliable methods to compute minor losses (Hooper’s 2-k method and Darby’s 3-k method).
Both approaches mentioned above, reveal that minor loss coefficients are strictly related to Reynolds number. Underestimating the effect of Re on minor loss can lead to heavy mistakes.

 

[katmar]

Using the Darby 3K method I get 7 psi for your fittings. But you have not defined your piping class so this can only be an estimate. The L/D method is much better for laminar flow than the fixed K method. In general L/D will over estimate minor losses, but I am surprised that it is out by this much. Perhaps it is because of your extremely low Reynolds number.

Relative to my 7 psi your L/D pressure drop is wrong by a factor of 2, while the fixed K method is out by a factor of 9. But both are out by around 7 psi in absolute terms.

Katmar Software
Engineering & Risk Analysis Software

http://katmarsoftware.com

 

[Superfly257]

Thanks for your help!

I was wondering if either of you could explain the Hooper 2K and Darby 3K method? Or if you could provide a link that might clear things up?

Thanks again,
-Superfly

 

[ione]

It’s kind of beating a dead horse

Both Hooper’s 2-k method and Darby’s 3-k method account for minor loss sensitivity to Reynolds number and scale effect.

For minor losses we can write:

h = Kf*V^2/2

Being
h = minor los
Kf = minor loss coefficient
V = fluid velocity in the pipe where the fitting is placed


Standing to 2-k method:

Kf = K1/Re + K ?*(1+ 1/IDinch)

Where:

Re = Reynolds number
ID = inner diameter of the pipe containing the fitting (in inch)
K1, K ? tabulated coefficient for each specific fitting


Standing to 3-k method:

Kf = K1/Re + Ki*(1+ Kd/D^0.3)

Where:

K1,Ki, Kd are tabulated for each specific fitting

The 3k-method is the most reliable method to predict minor pressure loss (not a surprise that respectful member katmar has used this approach).
I warmly suggest you to look for "Chemical Engineering Fluidmechanics" Second Edition written by Ron Darby.

To start you can give a read to

http://www.cheresources.com/eqlength.shtml

and to the thread below

http://www.eng-tips.com/viewthread.cfm?qid=173164

 

[ione]

The right formulation is obviously:

h = Kf*V^2/(2g)

with g = gravity acceleration

 

[Superfly257]

Thanks everyone for your valuable information!

I've obtained a copy of "Chemical Engineering Fluidmechanics" Second Edition by Ron Darby and have been looking through it. It has been, and will be very useful.

However, I have tried to calculate the minor pressure loss using the Darby 3-K method, and I am coming up with a different answer than Katmar, and I was hoping I could get some clarification so I know if I'm doing it right.

For the Darby's book my K values, and the subsequent calculation of kf,w ere taken as follows:

K1 Ki Kd kf
5 Run through, threaded Tees = 200 0.091 4 3.877
4 standard, threaded 90 Elbow= 800 0.14 4 14.310
2 Fully open Ball Valves = 300 0.017 4 5.176

Next I added all the Kf values, Kft, and used this equation to solve for pressure loss:

Pressure loss = kft*Density*V^2/2

After converting the answer to PSI, I had only 1.698 PSI pressure loss.

I would very much like to know how to do this right, so I would be very grateful for any light you could shed on my mishaps!

Thanks again,
-Superfly

PS: All the values I used in my calculation, like diameter, flow rate, Reynolds number, etc, are the same as in the spreadsheet I uploaded earlier, none of that has changed.

 

[stanier]

The variation in inside diameter will have more impact than minor losses. Seamless carbon steel has a 12.5% allowance on wall thickness. All thermoplastics have ~10% wall thickness variation. GRp is a basket case. Cement lined steel and ductile iron has tolerance on the metal and the cement lining.

"Sharing knowledge is the way to immortality"
His Holiness the Dalai Lama.

http://waterhammer.hopout.com.au/

 

[katmar]

It looks like you have neglected to multiply the k_f's for each type of fitting by the number of fittings.

If you correct this you will get a slightly lower number than I reported earlier because I assumed reduced bore ball valves.

Stanier brings up a valid point, but this effect will not swamp the effect of the minor losses in every case. It is still worth calculating the impact of the fittings, and indeed the effect of the change in pipe ID due to wall thickness tolerance (in both directions). If the wall of a Sched 40 3/4" pipe varies by 12.5% it will only have a <2% impact on the ID and in this particular example this results in less pressure drop change than that caused by the fittings.

Katmar Software
Engineering & Risk Analysis Software

http://katmarsoftware.com

 

[ione]

Superfly257,

I’ve got 6.97 psi with your data (Kf tot = 27.944), that is what katmar's got.

h = Kf tot * V^2/(2g) = 27.944* 11.296 (ft^2/s^2)/[2*32.185 (ft/s^2)] = 16.088 ft H2O = 6.97 psi

I think the point risen by stanier has to be taken into account for turbulent regime, but with such a low Re the impact of wall thickness allowance is of secondary importance (as noted above)

 

[ione]

Yup, I’ve entered completely wrong numbers

Kf tot = 92.358

So

h = 18.9 ft H2O = 8.2 psi

 

[Superfly257]

Thanks everyone for all this valuable advice and input!

I've finally got an answer that squares with Katmar's.
I'm still very new to this whole game, so I truly appreciate all the help you've been giving me, and now (after all your posting, and reading through Darby's texts) I can finally feel comfortable that I'm starting to understand all this, and that I'm doing the calculations right!

There are a few things I would just like to clear up though:
When dealing with low Reynold's number, laminar flow...
Using Equivalent Lengths will typically be a conservative estimate, as it tends to overestimate the minor pressure losses?
Using Crane's coefficient method will tend to under estimate the friction losses?
And using the Darby 3-K method is generally accepted as the most accurate for low Reynold's number flows, and will be closest to the real world pressure loss of the system?

Also, is there any problem with using the 3-K method at turbulent Reynold's numbers? I know that Crane's method is a good approximation when dealing with turbulent flow, but is the 3-K method still considered any more or less accurate?

Thanks,
-Superfly

 

[ione]

The strength of 3k method is that, whilst being the most reliable method in accounting for Re impact in the laminar flow regime, it also produces results that are strictly in agreement with those obtained with Crane method when dealing with turbulent regime.